Automatic Control Systems Solutions Manual Kuo Golnaragh by Benjamin Kuo PDF

By Benjamin Kuo

Similar mathematics books

New PDF release: Elements de Mathematique. Algebre commutative. Chapitres 5a7

Les ? ‰l? ©ments de math? ©matique de Nicolas Bourbaki ont pour objet une pr? ©sentation rigoureuse, syst? ©matique et sans pr? ©requis des math? ©matiques depuis leurs fondements. Ce deuxi? ?me quantity du Livre d Alg? ?bre commutative, septi? ?me Livre du trait? ©, introduit deux notions fondamentales en alg?

This can be the second one printing of the ebook first released in 1988. the 1st 4 chapters of the amount are in keeping with lectures given by means of Stroock at MIT in 1987. They shape an creation to the fundamental rules of the speculation of enormous deviations and make an appropriate package deal on which to base a semester-length path for complicated graduate scholars with a robust heritage in research and a few chance concept.

Additional info for Automatic Control Systems Solutions Manual Kuo Golnaragh

Sample text

4 !   cos t = − sin t sin t  cos t  (2) Inverse Laplace transform: −1 1  s 1  s −1  Φ ( s ) = ( sI − A ) = 1 s  = s 2 + 1  −1 s  −1 φ (t ) =  cos t  − sin t sin t  cos t  (b) A=  −1 0   0 −2  A = 2 1 0   0 4  A = 3  −1 0   0 − 8  A = 4  −1 0   0 16  (1) Infinite series expansion: 2 3 4 t t t   1 − t + − + +L 0   1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L =   2 3 2! 4t 8t   0 1 − 2t + − +L   2! 3! e −t = 0   e  0 −2 t (2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1 s + 1  0  1   s +1 = s + 2   0     1  s + 2  0 −1 0  e− t  −2 t  e  0 φ (t ) =  0 (c) A= 0 1  1 0  A = 2 1 0  0 1  A = 3 0 1  1 0  A = 4 1 0  0 1  (1) Infinite series expansion: 3 5 t t  t2 t4  1+ + +L t+ + L   1 2 2  e− t + et 2!

2! 4 !   cos t = − sin t sin t  cos t  (2) Inverse Laplace transform: −1 1  s 1  s −1  Φ ( s ) = ( sI − A ) = 1 s  = s 2 + 1  −1 s  −1 φ (t ) =  cos t  − sin t sin t  cos t  (b) A=  −1 0   0 −2  A = 2 1 0   0 4  A = 3  −1 0   0 − 8  A = 4  −1 0   0 16  (1) Infinite series expansion: 2 3 4 t t t   1 − t + − + +L 0   1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L =   2 3 2! 4t 8t   0 1 − 2t + − +L   2! 3! e −t = 0   e  0 −2 t (2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1 s + 1  0  1   s +1 = s + 2   0     1  s + 2  0 −1 0  e− t  −2 t  e  0 φ (t ) =  0 (c) A= 0 1  1 0  A = 2 1 0  0 1  A = 3 0 1  1 0  A = 4 1 0  0 1  (1) Infinite series expansion: 3 5 t t  t2 t4  1+ + +L t+ + L   1 2 2  e− t + et 2!

The last column of Thus, the last column of adj (s I − A ) adj (s I − A ) B 50 is 1 adj (s I − A ) is is obtained from the cofactors of s s 2 L s n −1 ' . 0 x 5-18 (a) State variables: x 1 = y, x 2 = dy , dt x 3 = d 2 y 2 dt x& ( t ) = Ax ( t ) + B r ( t ) State equ ations: 0 1 0 A= 0 0 1    −1 −3 −3   0 B =  0   1  (b) State transition matrix: s2 + 3 s + 3 s + 3 1   s −1 0  1  −1  2 Φ ( s ) = ( sI − A ) =  0 s −1  = −1 s + 3s s     ∆ (s ) 2  − s  1 3 s + 3  −3 s − 1 s   1 1 1 1 2 1  + + +   2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3   s + 1 ( s + 1) ( s + 1)   −1 1 1 2 s = + −  2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3     2 −s −3 2 s   +  ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3  −1 ∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1) (1 + t + t 2 / 2 ) e− t  2 −t φ ( t) =  −t e / 2  ( −t + t 2 / 2 ) e −t (t + t ) e (1 + t − t ) e −t 2 2 2 t e 3  ( t − t / 2 ) e  (1 − 2t + t 2 / 2 ) e −t  2 −t t e /2 −t −t 2 −t (d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3 − 1, −1, −1 Eigenvalues: 5-19 (a) State variables: 2 x 1 = y, x 2 = dy dt State equations:  dx1 (t )   dt   0 1   x1 ( t)   0   =   +   r ( t)  dx2 ( t)   − 1 − 2   x2 ( t)   1   dt  State transition matrix:  s+ 2  ( s + 1) 2  s −1  −1  Φ ( s ) = ( sI − A ) =  1 s + 2  =  − 1  ( s + 1) 2  −1  ( s + 1)  s  ( s + 1)2  1 2 51  (1 + t ) e−t φ (t ) =   − te −t   (1 − t ) e  te −t −t Characteristic equation: = y, −y= x (b) State variables: x 1 ∆ (s ) = ( s +1) = 0 2 x 2 = y dy + dt State equations: dx dt 1 = dy dt = x 2 dx − x1 2 2 dt d = 2 y dt dy + 2 dt = −y − dy +r = −x2 +r dt  dx1   dt   −1 2   x1  0   =   + r  dx2   0 −1  x2  1   dt  State transition matrix:  1 s + 1 −2   s + 1 Φ (s ) =  =  0 s + 1   0  ( s + 1)  1   s +1  −2 −1 2 (c) Characteristic equation: ∆ (s ) = ( s +1) = 0 φ (t ) = e −t 0  − te e −t −t    2 5-20 (a) State transition matrix: ω  s − σ sI − A =  −ω s − σ  ( sI − A ) −1 which is the same as in part (a).